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Proof step 17 from proof countFun-last-01 (previous step)

i ≤ i2, i ≤ i2 + 1, i1 + i = i2 + 1, 0 ≤ i2, i ≤ i2, 0 ≤ i, 0 ≤ i1, ∀ j0, i0, intfun. i→n(j0 - i0) < i→n(i1) → true → i0 ≤ j0 ∧ 0 ≤ i0 → countFun(i0, j0, intfun) = countFun(i0, j0 - 1, intfun) + intfun(j0) ' ⊦ intfun(i) ' + countFun(i + 1, i2 + 1, intfun) = (intfun(i) ' + countFun(i + 1, i2, intfun)) + intfun(i2 + 1) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
i1 + i = i2 + 1, 0 ≤ i2, i ≤ i2, 0 ≤ i, 0 ≤ i1, ∀ j0, i0, intfun. i→n(j0 - i0) < i→n(i1) → true → i0 ≤ j0 ∧ 0 ≤ i0 → countFun(i0, j0, intfun) = countFun(i0, j0 - 1, intfun) + intfun(j0) ' ⊦ countFun(i + 1, i2 + 1, intfun) = countFun(i + 1, i2, intfun) + intfun(i2 + 1) '

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%le-add-03: ⊦ i ≤ i0 ∧ j0 ≥ 0 → i ≤ i0 + j0; from specification int-abs%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%app-01: ⊦ ints + ints0 = ints + ints1 ↔ ints0 = ints1; from specification list[intlista]